$x=\sqrt{A^{2}+B}\left[\frac{A}{A^{2}+B^{2}} \sin \omega t+\frac{B}{\sqrt{A^{2}+B^{2}}} \cos \omega t\right]$
$\mathrm{x}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}[\cos \phi \sin \omega \mathrm{t}+\sin \phi \cos \omega \mathrm{t}]$
$\mathrm{x}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}} \sin (\omega \mathrm{t}+\phi)$
$Y = A \sin (\pi t +\phi)$, where time is measured in $second$.
The length of pendulum is .............$cm$
Where $A$ and $K$ are positive constants
